Problem: Consider the polar curve $r=\sin^2\left(2\theta\right) $. What is the equation of the tangent line to the curve $r$ at $\theta=\dfrac{\pi}{4}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-\dfrac{\sqrt{2}}{2}=-\left(x-\dfrac{\sqrt{2}}{2}\right)$ (Choice B) B $y-\dfrac{\sqrt{2}}{2}=-\dfrac{\sqrt{2}}{2}\left(x-\dfrac{\sqrt{2}}{2}\right)$ (Choice C) C $y+1=-\dfrac{\sqrt{2}}{2}\left(x-\dfrac{\pi}{4}\right)$ (Choice D) D $y-1=-\left(x-\dfrac{\pi}{4}\right)$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Then we can use the point-slope form to complete the equation for the tangent line through $\theta=\dfrac{\pi}{4}$. For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={\sin^2(2\theta)}\cos(\theta) \\\\ y&={\sin^2(2\theta)} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{2\sin(4\theta)\sin(\theta)+\sin^2(2\theta)\cos(\theta)}{2\sin(4\theta)\cos(\theta)-\sin^2(2\theta)\sin(\theta)} \end{aligned}$ Note: Because $\sin(2\theta)\neq0$ when $\theta=\dfrac{\pi}{4}$, we can divide throughout by $\sin(2\theta)$ to simplify. Evaluating $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{4}}$ gives us the slope of our tangent line. $\begin{aligned} &\phantom{=}{\left. \dfrac{dy}{dx}\right| _{\theta=\tfrac{\pi}{3}}} \\\\ &=\dfrac{2\sin\left(4\left({\dfrac{\pi}{4}}\right)\right)\sin\left({\dfrac{\pi}{4}}\right)+\sin^2\left(2\left({\dfrac{\pi}{4}}\right)\right)\cos\left({\dfrac{\pi}{4}}\right)}{2\sin\left(4\left({\dfrac{\pi}{4}}\right)\right)\cos\left({\dfrac{\pi}{4}}\right)-\sin^2\left(2\left({\dfrac{\pi}{4}}\right)\right)\sin\left({\dfrac{\pi}{4}}\right)} \\\\ &=\dfrac{2\sin(\pi)\sin\left(\dfrac{\pi}{4}\right)+\sin^2\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{\pi}{4}\right)}{2\sin(\pi)\cos\left(\dfrac{\pi}{4}\right)-\sin^2\left(\dfrac{\pi}{2}\right)\sin\left(\dfrac{\pi}{4}\right)} \\\\ &=\dfrac{2(0)\left(\dfrac{\sqrt 2}{2}\right)+(1)^2\left(\dfrac{\sqrt 2}{2}\right)}{2(0)\left(\dfrac{\sqrt 2}{2}\right)-(1)^2\left(\dfrac{\sqrt 2}{2}\right)} \\\\ &={-1} \end{aligned}$ We now find $x$ and $y$ at the point $\theta=\dfrac{\pi}{4}$. $\begin{aligned} {x\left({\dfrac{\pi}{4}}\right)}&=\sin^2\left(2\left({\dfrac{\pi}{4}}\right)\right)\cos\left({\dfrac{\pi}{4}}\right) \\\\ &=\sin^2\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{\pi}{4}\right) \\\\ &={\dfrac{\sqrt{2}}{2}} \\\\ \\\\ y\left({\dfrac{\pi}{4}}\right)}&=\sin^2\left(2\left({\dfrac{\pi}{4}}\right)\right)\sin\left({\dfrac{\pi}{4}}\right) \\\\ &=\sin^2\left(\dfrac{\pi}{2}\right)\sin\left(\dfrac{\pi}{4}\right) \\\\ &=\dfrac{\sqrt{2}}{2}} \end{aligned}$ Therefore the equation of our tangent line is: $\begin{aligned} y-\dfrac{\sqrt{2}}{2}}&={(-1)}\left(x-{\dfrac{\sqrt{2}}{2}}\right) \\\\ y-\dfrac{\sqrt{2}}{2}&=-\left(x-\dfrac{\sqrt{2}}{2}\right) \end{aligned}$ The graph of the tangent is shown.